The definition of the product, K×LK\times L, of simplicial sets KK and LL gives that (K×L)k(K\times L)_k is defined to be Kk×LkK_k\times L_k with face and degeneracy maps defined ‘componentwise’ so, for instance, di(x,y)=(dix,diy)d_i(x,y) = (d_i x,d_i y). This means that a simplex(sαx,sβy)(s_\alpha x, s_\beta y) can be non-degenerate even though its two components are degenerate. Of course, this happens exactly when α∩β=∅\alpha \cap \beta = \emptyset. (Here we mean by sαs_\alpha the composite, in order, of the sis_i for i∈αi \in \alpha.)

In particular, take K=Δ[p]K = \Delta[p] and L=Δ[q]L= \Delta[q] with xp∈Δ[p]px_p \in \Delta[p]_p, yq∈Δ[q]qy_q \in \Delta[q]_q, the ‘identity’ simplices that generate them. (We will use an ad hoc notation here, since the more obvious, ιp\iota_p, and ιq\iota_q, would not distinguish the two simplices sufficiently in some formulae.) There are, then, non-degenerate simplices of dimension p+qp+q, but none of higher dimension in Δ[p]×Δ[q]\Delta[p]\times \Delta[q]. These are of the form (sαxp,sβyq)(s_\alpha x_p,s_\beta y_q), where #α=q\#\alpha = q, #β=p\#\beta = p.

Non-degenerate simplices in a product

If we represent a vertex of a product by a ‘column vector’ rather than the more usual ‘row vector’ for the moment, then any non-degenerate (p+q)(p+q)-simplex of Δ[p]×Δ[q]\Delta[p]\times \Delta[q] can be represented in the form of an ordered list of vertices,

with p+q+1p + q + 1 columns. The top row, which is a simplex in Δ[p]\Delta[p], changes at exactly those positions at which the bottom row repeats. The top row gives a degenerate p+qp+q simplex of Δ[p]\Delta[p], whilst the bottom row one of Δ[q]\Delta[q], i.e., the array gives (sαxp,sβyq)(s_\alpha x_p, s_\beta y_q) for some (α,β)(\alpha, \beta) as above.

Illustrative example

The usual illustrative example is of the three 3-simplices of Δ[2]×Δ[1]\Delta[2]\times \Delta[1]:

Simplices of the product and partitions

Each such simplex yields a partition of {0,…,p+q−1}\{0, \ldots, p+q-1\} into two disjoint sets, μ1<…<μp\mu_1\lt\ldots \lt\mu_p, and ν1<…<νq\nu_1 \lt \ldots \lt \nu_q, and vice versa. Suppose that we have an array, as above, written

with 0=i0≤i1≤…≤ip+q=p0= i_0 \leq i_1 \leq \ldots \leq i_{p+q}= p, then if ik=ik+1i_k = i_{k+1}, we put kk into the second set, otherwise kk is put in the first set. This, of course, leads to an operation that preserves order. For instance, in the above example 2., the ii-sequence is (0,1,1,2)(0,1,1,2), so there is the single repeat with k=1k = 1, and ν={1}\nu = \{1\}.

We likewise require 0=j0≤j1≤…≤jp+q=p0= j_0 \leq j_1 \leq \ldots \leq j_{p+q}= p, and put kk into the first set if jk=jk+1j_k = j_{k+1}. For the example, we have the jj-sequence is (0,0,1,1)(0,0,1,1), so μ={0,2}\mu = \{ 0,2\}. Of course, from the partition you can get the sequences and conversely. The attentive reader will, of course, have noted that, for example 2., the α\alpha, we specified was exactly the ν\nu, and the β\beta was the μ\mu. This is general with the simplex corresponding to a partition, (μ,ν)(\mu,\nu), being given by (sνq…sν1xp,sμp…sμ1yq)(s_{\nu_q}\ldots s_{\nu_1}x_p,s_{\mu_p}\ldots s_{\mu_1}y_q).

Each such partition defines a permutation of {0,…,p+q−1}\{0,\ldots, p+q-1 \}. Let us write ι0:{0,…,q−1}→{0,…,p+q−1}\iota_0 : \{ 0, \ldots, q-1\}\to \{0, \ldots, p+q-1\} for the order preserving function ι0(r)=p+r\iota_0 (r)= p+r, whilst ι1:{0,…,p−1}→{0,…,p+q−1}\iota_1 : \{ 0, \ldots, p-1\}\to \{0, \ldots, p+q-1\} will denote the inclusion, so ι1(r)=r\iota_1(r) = r. There will be a permutation σ\sigma of {0,…,p+q−1}\{0,\ldots, p+q-1 \} such that σι0(r)=νr+1\sigma \iota_0(r) = \nu_{r+1} and σι1(r)=μr+1\sigma\iota_1(r) = \mu_{r+1}. This means that the permutation looks like

We can thus assign a sign, sgn(σ)sgn(\sigma), to each such shuffle, namely the sign of the corresponding permutation.

For our standard examples, we have : 1) σ\sigma is the identity, 2) σ=(012021)\sigma = \left(\begin{array}{ccc}0&1&2\\0&2&1\end{array}\right), i.e. the transposition exchanging 1 and 2, and for 3) σ=(012120)\sigma = \left(\begin{array}{ccc}0&1&2\\1&2&0\end{array}\right), a 3-cycle. We thus have, in this case, the signs are +1, -1, and + 1, respectively.

Paths in the product

A final useful interpretation of (p,q)(p,q) shuffles is of ascending paths through a pp by qq integer lattice from (0,0)(0,0) to (p,q)(p,q). This is quite well illustrated by our example. The vertices are the integer pairs, (i,j)(i,j) with 0≤i≤p0\leq i\leq p and 0≤j≤q0\leq j\leq q, so in our case we get

The path corresponding to a (p,q)(p,q)-shuffle just follows the list of (transposed pairs of) vertices, so, for instance, 2) goes

The poset of (p,q)(p,q)-shuffles

Any pair (p,q)(p,q) yields a poset relating the various (p,q)(p,q)-shuffles.

Easy examples

For the sake of the will give some elementary examples. (This section can be skipped if you want to move on to the more theoretical analysis of these posets.)

Our (2,1)(2,1)-example is really too simple and small to illustrate this well, but the two cases (3,1)(3,1) and (2,2)(2,2) do a much better job, but, even so, we first look at the (2,1) example:

(This Hasse diagram has been laid out horizontally to save space. The bottom is to the left. The vertex (0<1)(0\lt 1) corresponds to the shuffle with μ1=0,μ1=1\mu_1 = 0, \mu_1 = 1, and so on.)

We need here to explain the partial order. We take the μ\mu-signature’ of the shuffle, that is, the ordered set μ1<…<μp\mu_1\lt\ldots \lt \mu_p. (Of course, this determines the ν\nu-signature as that is the complement of μ\mu.)

Definition

Given two (p,q)(p,q)-shuffles, represented by (μ,ν)(\mu, \nu) and (μ′,ν′)(\mu\prime,\nu\prime), we set

(μ,ν)≤(μ′,ν′)(\mu, \nu) \leq (\mu\prime,\nu\prime)

if, for each ii in the range 1≤i≤p1\leq i\leq p, μi≤μi′.\mu_i \leq \mu_i\prime. We refer to this poset as (Shuff(p,q),≤)(Shuff(p,q),\leq).

Going to (3,1)(3,1), the fact that q=1q = 1 will mean that the poset is linear:

This is general:

Lemma

If p=1p = 1 or q=1q = 1, then (Shuff(p,q),≤)(Shuff(p,q),\leq) is a linear poset.

Proof

If p=1p = 1, μ=(μ1)\mu = (\mu_1) is a singleton, and the poset will be:

For q=1q = 1, ν=(ν1)\nu = (\nu_1), and the poset is

where at each stage one misses out the single ν\nu-value.

In all cases, each position is obtained from the one immediately to its ‘left’ by increasing one value, yet remaining with a shuffle. This is more clearly seen in the (2,2) example, which is no longer linear. First we display the grid in which things are happening.

We can then look at the shuffle poset, noting again that it is not linear:

The left hand shuffle, labelled (0<1)(0\lt1), corresponds to (0122200012)\left(\begin{array}{ccccc}0&1&2&2&2\\0&0&0&1&2\end{array}\right), so gives the path along the bottom of the square and then up the right hand side.

The first change goes to (0<2)(0\lt 2) and gives a path with 2 steps,

and corresponds to the shuffle, (0112200112)\left(\begin{array}{ccccc}0&1&1&2&2\\0&0&1&1&2\end{array}\right), so at this position, (0<2)(0\lt2), there is a choice, either increase 0 by 1 to get (1<2)(1\lt2) or increase 2 by 1 to get (0<3)(0\lt3). In the first case, we get the path shuffle

and the shuffle, (0012201112)\left(\begin{array}{ccccc}0&0&1&2&2\\0&1&1&1&2\end{array}\right), so at this position, there is a choice, either increase 0 by 1 to get (1<2)(1 \lt 2):

or increase 2 to 3 to get (0<3)(0 \lt 3), the shuffle: (0111200122)\left(\begin{array}{ccccc}0&1&1&1&2\\0&0&1&2&2\end{array}\right), and the obvious path up the middle of the square:

From (1<2)(1 \lt 2) or (0<3)(0\lt3), there is only on way to go, namely to (1<3)(1 \lt 3) and a 2-step path (left to you), and, finally it is just one step from (1<3)(1 \lt 3) to (2<3)(2 \lt 3) and the other extremal path.

Summary

Each path corresponds to a simplex of maximal dimension in the product. The poset encodes the simplest relationships between those paths with the links in the Hasse diagram corresponding to simple changes in the path, and adjacency of the two simplices in the product, but note that the poset is usually not linear.

The Anti-Lex order

There is a useful total order on Shuff(p,q)Shuff(p,q) related to the partial order and which is useful when checking cancellation of terms in non-commutative contexts. This is the anti-lexicographic order.

as, at each position, there is only one transposition that can be applied. The anti-lex total order will correspond exactly to this if p−1p-1 is odd, but, if p−1p-1 is even, it will reverse the order on the last p−1p-1 positions giving

There are various points to note. Firstly that if pp is even, the permutation corresponding to (1,…,p−1)(1, \ldots , p-1) is odd. Secondly, the geometric picture is simply a prism, Δ[p]×Δ[1]\Delta[p]\times \Delta[1], and we can easily interpret the above order as a filling scheme for the simplices of (Δ[p]×Δ[1])×Δ[1](\Delta[p]\times \Delta[1])\times \Delta[1], starting with the empty shell of